1.

2C4H10 + 13O2 → 8CO2 + 10H2O This is the equation of ignition of cooking gas butane.Calculate the volume of CO2 in STP during the complete ignition of 14 kg of cooking gas.

Answer»

Mass of Butane (C4H10) = 14 kg = 1400g

GMM of C4H10 = 4 × 12 + 10 × 1 = 58 g

∴ Number of moles in molecules = \(\frac{1400}{58}\) = 241.38

Amount of CO2 when 2 moles of C4H10 ignites = 8 moles of C4H10 ignites = \(\frac{8}{2}\)× 965.52 moles

∴ Volume of CO2 formed in STP 

= 965.52 × 22.4 L = 21627.65 L



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